I am trying to solve Exercise 6.5 part 4 in James Humphreys' Introduction to Lie Algebras and Representation Theory. I added the (homework) tag because my question is about an exercise, but this is not really homework, I am really looking for a proof of the statement. Anyway, this is the exercise:

Exercise 6.5. A Lie algebra $L$ for which $\DeclareMathOperator{\Rad}{Rad}\Rad(L)=Z(L)$ is called reductive. (Examples: $L$ abelian, $L$ semisimple, $L=\newcommand{\gl}{\mathfrak{gl}}\gl(n,\newcommand{\HF}{\mathbb F}\HF)$.)

  1. If $L$ is reductive, then $L$ is a completely reducible $\DeclareMathOperator{\ad}{ad}\ad(L)$-module. [If $\ad(L)\ne 0$, use Weyl's Theorem.] In particular, $L$ is the direct sum of $Z(L)$ and $[LL]$, with $[LL]$ semisimple.
  2. If $L$ is a classical linear Lie algebra (1.2), then $L$ is semisimple
  3. If $L$ is a completely reducible $\ad(L)$-module, then $L$ is reductive.
  4. If $L$ is reductive, then all finite dimensional representations of $L$ in which $Z(L)$ is represented by semisimple endomorphisms are completely reducible.

What I tried so far: Write $R=\Rad(L)$ for the radical of $L$ and $Z=Z(L)$ for the center of $L$.

Let $\rho:L\to\gl(V)$ be a finite-dimensional representation of $L$ such that $Z$ acts by semisimple endomorphisms of $V$. Let $K:=[LL]$, so $L=K\oplus Z$ by part (1). For any $x\in L$ and $y\in Z$, $$ 0=\rho[xy]=[\rho(x)\rho(y)]=\rho(x)\rho(y)-\rho(y)\rho(x), $$ so $\rho(x)$ and $\rho(y)$ commute. In particular, all elements of $\rho(Z)$ commute, so we may simultaneously diagonalize all of them. In other words, there is a basis $v_1,\ldots,v_r$ of $V$ such that each $v_i$ is an eigenvector for each endomorphism in $\rho(Z)$.

Hence, the representation $\rho|_Z:Z\to\gl(V)$ completely decomposes into these eigenspaces. On the other hand, $K$ is semisimple, so $\rho|_K:K\to \gl(V)$ also completely decomposes by Weyl's Theorem. Let now $W$ be any $L$-submodule of $V$. Then, $W$ is, in particular, both a $K$ and a $Z$-module by restricting. Hence, $W$ has a $K$-complement $U$ in $V$. We want to show that $U$ is $Z$-invariant.

For every $i$, there exists some $\lambda\in\HF$ such that $$ y.(x.v_i)=x.(y.v_i)=x.(\lambda v_i)=\lambda(x.v_i), $$ so $x.v_i$ is an eigenvector of $\rho(y)$. Hence, any $x\in L$ sends eigenvectors to eigenvectors.

Here, I am stuck. I do not see how to conclude from the above that $U$ is invariant under the action of $Z$ - all bearing in mind that maybe this whole approach is useless and I need a different one.

  • $\begingroup$ The conclusion of part 1 of exercise 6.5 is missing the possibility that $[LL]=0$. $\endgroup$
    – JDZ
    Jul 13, 2018 at 0:53
  • $\begingroup$ @JDZ: When $[LL]=0$, then $Z(L)=L$ and $\operatorname{ad}(L)=0$, and all this is trivially true. So nothing is missing in the conclusion of part 1 here. $\endgroup$
    – ladisch
    Jul 13, 2018 at 11:11

1 Answer 1


The problem with your approach is that $W$ may have more than one $K$-complement (think of the case $K=0$) and not every complement may be $Z$-invariant. So it's better to do it the other way round: As you said, there is a basis of $V$ such that each element of this basis is an eigenvector for each $\rho(z)$. Let $v$ be such an eigenvector, so that for each $z\in Z$ there is $\alpha(z)\in \mathbb{F}$ such that $z.v= \alpha(z)v$. The map $\alpha\colon Z\to \mathbb{F}$ is linear. Set $$ V_{\alpha} = \{v\in V \mid z.v = \alpha(z)v \quad\text{for all } z\in Z \}.$$ Since $\rho(Z)$ is simultaneously diagonizable, we have that $$ V= \bigoplus_{\alpha} V_{\alpha}, $$ where $\alpha$ runs over linear maps $\alpha\colon Z\to \mathbb{F}$. (Of course, $V_{\alpha}=0$ for all but finitely many $\alpha$'s.)
The $V_{\alpha}$'s are $L$-invariant: for $v\in V_{\alpha}$, $z\in Z$ and $x\in L$ we have $$ z.(x.v) = x.(z.v)+[z,x].v = x.(z.v) = x.(\alpha(z)v)=\alpha(z)(x.v), $$ so $x.v\in V_{\alpha}$.
Finally, each $V_{\alpha}$ decomposes as a direct sum of irreducible $K$-submodules. Submodules of $V_{\alpha}$ are $Z$-invariant, since $Z$ acts as scalars on $V_{\alpha}$. This gives the desired decomposition of $V$ into irreducible $L$-submodules.

  • $\begingroup$ That's perfect! $\endgroup$
    – user38451
    Aug 30, 2012 at 9:35
  • $\begingroup$ What if $[LL]=0$? We can't decompose the $V_{\alpha}$ by Weyl's theorem in this case. $\endgroup$
    – JDZ
    Jul 13, 2018 at 0:56
  • $\begingroup$ @JDZ: When $[LL]=0$, every decomposition into $1$-dim subspaces of $V_{\alpha}$ will do. $\endgroup$
    – ladisch
    Jul 13, 2018 at 11:12

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